Question: $d(n) = \dfrac{5}{16} \left(2\right)^{n - 1}$ What is the $5^\text{th}$ term in the sequence?
This is an explicit formula. All we have to do is plug $n=5$ in the formula to find the $5^\text{th}$ term. $\begin{aligned} d({5}) &=\dfrac{5}{16}(2)^{{5} - 1} \\\\ &= 5 \end{aligned}$ The $5^\text{th}$ term is $5$.